FP3 June 2011 Q8
8. The hyperbola \(H\) has equation \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]
The line \(l_1\) is the tangent to \(H\) at the point \((a\cosh\theta, b\sinh\theta),\ \theta \neq 0\).
Given that \(l_1\) meets the \(x\)-axis at the point \(P\),
The line \(l_2\) is the tangent to \(H\) at the point \((a, 0)\).
Given that \(l_1\) and \(l_2\) meet at the point \(Q\),
| Scheme | Marks |
|---|---|
| Uses \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}} = \dfrac{b\cosh\theta}{a\sinh\theta}\) or \(\dfrac{2x}{a^2} - \dfrac{2yy^{\prime}}{b^2} = 0 \to y^{\prime} = \dfrac{xb^2}{ya^2} = \dfrac{b\cosh\theta}{a\sinh\theta}\) | M1 A1 |
| So \(y - b\sinh\theta = \dfrac{b\cosh\theta}{a\sinh\theta}(x - a\cosh\theta)\) | M1 |
| \(\therefore ab(\cosh^2\theta - \sinh^2\theta) = xb\cosh\theta - ya\sinh\theta\) and as \((\cosh^2\theta - \sinh^2\theta) = 1\) \(xb\cosh\theta - ya\sinh\theta = ab\) * | A1cso |
| (4) |
Notes
1M1 Finding gradient in terms of \(\theta\)
1A1 CAO
2M1 Finding equation of tangent
2A1 CSO (answer given) look for \(\pm(\cosh^2\theta - \sinh^2\theta)\)
| Scheme | Marks |
|---|---|
| \(P\) is the point \(\left(\dfrac{a}{\cosh\theta}, 0\right)\) | M1 A1 |
| (2) |
Notes
M1 Putting \(y = 0\) into their tangent
A1ft P found, ft for their tangent o.e.
| Scheme | Marks |
|---|---|
| \(l_2\) has equation \(x = a\) and meets \(l_1\) at \(Q\left(a, \dfrac{b(\cosh\theta - 1)}{\sinh\theta}\right)\) | M1 A1 |
| (2) |
Notes
M1 Putting \(x = a\) into their tangent.
A1 CAO Q found o.e.
| Scheme | Marks |
|---|---|
| Alt 1 The mid point of \(PQ\) is given by \(X = \dfrac{a(\cosh\theta + 1)}{2\cosh\theta},\quad Y = \dfrac{b(\cosh\theta - 1)}{2\sinh\theta}\) | 1M1 A1ft |
| \(4Y^2 + b^2 = b^2\left(\dfrac{\cosh^2\theta + 1 - 2\cosh\theta + \sinh^2\theta}{\sinh^2\theta}\right)\) | 2M1 |
| \(= b^2\left(\dfrac{2\cosh^2\theta - 2\cosh\theta}{\sinh^2\theta}\right)\) | 3M1 |
| \(X(4Y^2 + b^2) = ab^2\left(\dfrac{(\cosh\theta + 1)(\cosh\theta - 1)2\cosh\theta}{2\cosh\theta\sinh^2\theta}\right)\) | 4M1 |
| Simplify fraction by using \(\cosh^2\theta - \sinh^2\theta = 1\) to give \(x(4y^2 + b^2) = ab^2\) * | A1cso |
| (6) | |
| (14 marks) |
Notes
(d) Alt 2
| Scheme | Marks |
|---|---|
| First line of solution as before | 1M1A1ft |
| \(4Y^2 + b^2 = b^2\left(\coth^2\theta + \mathrm{cosech}^2\theta - 2\coth\theta\,\mathrm{cosech}\,\theta + 1\right)\) | 2M1 |
| \(= b^2\left(2\coth^2\theta - 2\coth\theta\,\mathrm{cosech}\,\theta\right)\) | 3M1 |
| \(X(4Y^2 + b^2) = ab^2\left(\coth\theta(\coth\theta - \mathrm{cosech}\,\theta)(1 + \mathrm{sech}\,\theta)\right)\) | 4M1 |
| Simplify expansion by using \(\coth^2\theta - \mathrm{cosech}^2\theta = 1\) to give \(x(4y^2 + b^2) = ab^2\) * | A1cso |
| (6) |
(d) Alt 3
| Scheme | Marks |
|---|---|
| \(X = \dfrac{a(\cosh\theta + 1)}{2\cosh\theta},\quad Y = \dfrac{b(\cosh\theta - 1)}{2\sinh\theta}\) As main scheme | 1M1 A1ft |
| \(\cosh\theta = \dfrac{a}{2x - a}\) \(\cosh\theta\) in terms of \(x\) | 2M1 |
| \(\sinh\theta = \dfrac{b(\cosh\theta - 1)}{2y} = \dfrac{b(a - x)}{(2x - a)y}\) \(\sinh\theta\) in terms of \(x\) and \(y\) | 3M1 |
| \(\left(\dfrac{a}{2x - a}\right)^2 - \left(\dfrac{b(a - x)}{(2x - a)y}\right)^2 = 1\) Using \(\cosh^2\theta - \sinh^2\theta = 1\) | 4M1 |
| Simplifies to give required equation \(\left[y^2 4x(a - x) = b^2(a - x)^2,\ x(4y^2 + b^2) = ab^2\right]\) | A1cso |
| (6) |
(d) Alt 4
| Scheme | Marks |
|---|---|
| \(X = \dfrac{a(\cosh\theta + 1)}{2\cosh\theta},\quad Y = \dfrac{b(\cosh\theta - 1)}{2\sinh\theta}\) As main scheme | 1M1 A1ft |
| \(\cosh\theta = \dfrac{a}{2x - a}\) \(\cosh\theta\) in terms of \(x\) | 2M1 |
| \(y^2 = \dfrac{b^2(\cosh\theta - 1)^2}{4(\cosh^2\theta - 1)} = \dfrac{b^2(\cosh\theta - 1)}{4(\cosh\theta + 1)}\) \(y^2\) in terms of \(\cosh\theta\) only | 3M1 |
| \(y^2 = \dfrac{b^2\left(\dfrac{2a - 2x}{2x - a}\right)}{4\left(\dfrac{2x}{2x - a}\right)}\) o.e Forms equation in \(x\) and \(y\) only | 4M1 |
| Simplifies to give required equation | A1 cso |
| (6) |
(corrected from the printed mark scheme: the printed line squares the bracket \(\left(\frac{2a - 2x}{2x - a}\right)\) in the numerator; with the denominator \(4\left(\frac{2x}{2x - a}\right)\) it is not squared, since \(y^2 = \frac{b^2(\cosh\theta - 1)}{4(\cosh\theta + 1)}\).)
(d) Alt 5
| Scheme | Marks |
|---|---|
| \(X = \dfrac{a(\cosh\theta + 1)}{2\cosh\theta},\quad Y = \dfrac{b(\cosh\theta - 1)}{2\sinh\theta}\) As main scheme | 1M1 A1ft |
| \(\cosh\theta = \dfrac{a}{2x - a}\) \(\cosh\theta\) in terms of \(x\) | 2M1 |
| \(y = \left(\dfrac{b(\cosh\theta - 1)}{2\sinh\theta}\right) = \left(\dfrac{b(\cosh\theta - 1)}{2\sqrt{\cosh^2\theta - 1}}\right)\) \(y\) in terms of \(\cosh\theta\) only | 3M1 |
| Eliminate \(\sqrt{\ }\) and forms equation in \(x\) and \(y\) | 4M1 |
| Simplifies to give required equation | A1cso |
Notes
(d) For Alt 1 and 2
1M1 Finding expressions, in terms of \(\sinh\theta\) and \(\cosh\theta\) but must be adding
1A1 Ft on their P and Q,
2M1 Finding \(4y^2 + b^2\)
3M1 Simplified, factorised, maximum of 2 terms per bracket
4M1 Finding \(x(4y^2 + b^2)\), completely factorised, maximum of 2 terms per bracket
2A1 CSO
(d) For Alts 3, 4 and 5
1M1 Finding expressions, in terms of \(\sinh\theta\) and \(\cosh\theta\) but must be adding
1A1 Ft on their P and Q
2M1 Getting \(\cosh\theta\) in terms of x
3M1 y or \(y^2\) in terms of \(\cosh\theta\) or \(\sinh\theta\) in terms of x and y
4M1 Getting equation in terms of x and y only. No square roots.
2A1 CSO